Chapter 5, Problem 121QRT

(a)

Interpretation Introduction

Interpretation:

The reason for the electron affinity of selenium being lower than that of bromine has to be explained using electron configurations.

Explanation of Solution

Electron affinity is the energy required to add an electron to the neutral atom.

  $\text{A}\text{+}{\text{e}}^{\text{-}}\stackrel{}{\to }{\text{A}}^{\text{-}}$

The electronic configuration of neutral selenium and selenium anion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Se}\text{(Z}\text{=}\text{34):}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{10}{\text{4p}}^{\text{4}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Se}}^{\text{-}}\text{:}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{10}{\text{4p}}^{\text{5}}\end{array}$

The electronic configuration of neutral bromine and bromine anion is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Br}\text{(Z}\text{=}\text{35):}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Br}}^{\text{-}}\text{:}\\ \left[{}_{\text{36}}\text{Kr}\right]\end{array}$

Addition of one electron to bromine gives a noble gas configuration.  Hence, it is very stable.  So, selenium has lower electron affinity than bromine.

(b)

Interpretation Introduction

Interpretation:

The reason for the first ionization energy of aluminum being lower than that of magnesium has to be explained using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral aluminum and aluminum cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Al}\text{(Z}\text{=}\text{13):}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{1}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Al}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}\end{array}$

The electronic configuration of neutral magnesium and magnesium cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Mg}\text{(Z}\text{=}\text{12):}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Mg}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{1}}\end{array}$

The magnesium atom has all subshells filled.  So, removing an electron requires more energy.  Hence, aluminum has lower first ionization energy than magnesium.

(c)

Interpretation Introduction

Interpretation:

The reason for the first ionization energy of sulfur being lower than that of phosphorus has to be explained using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral sulfur and sulfur cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{S}\text{(Z}\text{=}\text{16):}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{S}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{3}}\end{array}$

The electronic configuration of neutral phosphorus and phosphorus cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{P}\text{(Z}\text{=}\text{15):}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{3}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{P}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{2}}\end{array}$

The phosphorus atom has half-filledp-subshell.  So, removing an electron requires more energy.  Hence, sulfur has lower first ionization energy than phosphorus.

(d)

Interpretation Introduction

Interpretation:

The reason for the first ionization energy of bromine being lower than that of chlorine has to be explained using electron configurations.

Explanation of Solution

The ionization energy is the energy required to remove the outermost electron in an atom.

  $\text{A}\stackrel{}{\to }{\text{A}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

The electronic configuration of neutral bromine and bromine cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Br}\text{(Z}\text{=}\text{35):}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Br}}^{\text{+}}\text{:}\\ \left[{}_{\text{18}}\text{Ar}\right]{\text{4s}}^{\text{2}}{\text{3d}}^{\text{10}}{\text{4p}}^{\text{4}}\end{array}$

The electronic configuration of neutral chlorine and chlorine cation is written as follows:

  $\begin{array}{l}\text{Electronic}\text{configuration}\text{of}\text{Cl}\text{(Z}\text{=}\text{17):}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{5}}\\ \\ \text{Electronic}\text{cofiguration}\text{of}{\text{Cl}}^{\text{+}}\text{:}\\ \left[{}_{\text{10}}\text{Ne}\right]{\text{3s}}^{\text{2}}{\text{3p}}^{\text{4}}\end{array}$

The electron removed from chlorine is from $\text{3p}$ orbital whereas the electron removed from bromine is from $\text{4p}$ orbital.  Hence, bromine has lower first ionization energy than chlorine.